Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.lang.Math.*;

class Main{

    static double[][] dp = new double[3001][1502];
    static double[] mData = new double[3001];    
    static int N;
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in).useLocale(Locale.US);

        N = sc.nextInt();
        for(int i = 0; i < N; i++){
            mData[i+1] = sc.nextDouble();
        }
        
        int Npola = N/2 + 1;

        dp[1][0] = 1;
        dp[1][1] = mData[1];
        for(int i = 2; i < Npola; i++){
            dp[1][i] = 0;
        }

        for(int i = 2; i <= N; i++){
            dp[i][0] = 1;            
            for(int j = 1; j < 1502; j++){
                dp[i][j] = ( mData[i]*dp[i-1][j-1] + (1-mData[i])*dp[i-1][j] ); 
                //System.out.println( dp[i][j] );
            }
        }
    /*
        for(int i = 1; i <= N; i++){
            for(int j = 0; j <= N; j++){
                System.out.print(dp[i][j] + "\t");
            }
            System.out.println();
        }
    */
        System.out.println( dp[N][N/2 + 1] );
        sc.close();
    }
}